Show that the points P
Question:

Show that the points P(2, 3, 5), Q(-4, 7, -7), R(-2, 1, -10) and S(4, -3, 2) are the vertices of a rectangle.

 

Solution:

To prove: Points P, Q, R, S forms rectangle.

Formula: The distance between two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{Z}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ is given by

$\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$

Here,

$\left(x_{1}, y_{1}, z_{1}\right)=(2,3,5)$

$\left(x_{2}, y_{2}, z_{2}\right)=(-4,7,-7)$

$\left(x_{3}, y_{3}, z_{3}\right)=(-2,1,-10)$

$\left(x_{4}, y_{4}, z_{4}\right)=(4,-3,2)$

Length $P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

$=\sqrt{(-4-2)^{2}+(7-3)^{2}+(-7-5)^{2}}$

$=\sqrt{(-6)^{2}+(4)^{2}+(-12)^{2}}$

$=\sqrt{36+16+144}$

$=\sqrt{196}$

Length $Q R=\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}+\left(z_{3}-z_{2}\right)^{2}}$

$=\sqrt{(-2+4)^{2}+(1-7)^{2}+(-10+7)^{2}}$

$=\sqrt{(2)^{2}+(-6)^{2}+(-3)^{2}}$

$=\sqrt{4+36+9}$

$=\sqrt{49}$

Length $R S=\sqrt{\left(x_{4}-x_{3}\right)^{2}+\left(y_{4}-y_{3}\right)^{2}+\left(z_{4}-z_{3}\right)^{2}}$

$=\sqrt{(4+2)^{2}+(-3-1)^{2}+(2+10)^{2}}$

$=\sqrt{(6)^{2}+(-4)^{2}+(12)^{2}}$

$=\sqrt{36+16+144}$

$=\sqrt{196}$

Length $\mathrm{PS}=\sqrt{\left(\mathrm{x}_{4}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{4}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{4}-\mathrm{z}_{1}\right)^{2}}$

$=\sqrt{(4-2)^{2}+(-3-3)^{2}+(2-5)^{2}}$

$=\sqrt{(2)^{2}+(-6)^{2}+(-3)^{2}}$

$=\sqrt{4+36+9}$

$-\sqrt{49}$

Length PR $=\sqrt{\left(\mathrm{x}_{3}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{3}-\mathrm{z}_{1}\right)^{2}}$

$=\sqrt{(-2-2)^{2}+(1-3)^{2}+(-10-5)^{2}}$

$=\sqrt{(-4)^{2}+(-2)^{2}+(-15)^{2}}$

$=\sqrt{16+4+225}$

$=\sqrt{245}$

Length $Q S=\sqrt{\left(x_{4}-x_{2}\right)^{2}+\left(y_{4}-y_{2}\right)^{2}+\left(z_{4}-z_{2}\right)^{2}}$

$=\sqrt{(4+4)^{2}+(-3-7)^{2}+(2+7)^{2}}$

$=\sqrt{(8)^{2}+(-10)^{2}+(9)^{2}}$

$=\sqrt{64+100+81}$

$=\sqrt{245}$

Here, $P Q=R S$ which are opposite sides of polygon.

$\mathrm{QR}=\mathrm{PS}$ which are opposite sides of polygon.

Also the diagonals $P R=Q S$.

Hence, the polygon is a rectangle.

 

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